Work, Energy, and Power Chapter 5

Work Work – the product of the magnitudes of the component of force (along the direction of displacement) and the displacement W = Fd Work is done only when components of a force are parallel to a displacement

Work (cont.) Work is done only when components of a force are parallel to a displacement

Work (cont.) Work is done only when components of a force are parallel to a displacement W = F d (cos θ) W net = F net d (cos θ)

Sample Problem 5A p. 169 How much work is done on a vacuum cleaner pulled 3.0 m by a force of 50.0 N at an angle of 30.0 o above the horizontal? Given: F = 50.0 N, d = 3.0 m, θ = 30.0 o Unknown: W Equation: W = F d (cos θ) Substitute: W = 50 N(3m) cos 30 Solution: W = 130 J

Practice 5A 1. A tugboat pulls a ship with a constant net horizontal force of 5.00 x 10 3 N and causes the ship to move through a harbor. How much work is done on the ship if it moves a distance of 3.00 km? (1.50 x 10 7 J) 2. A weight lifter lifts a set of weights a vertical distance of 2.00 m. If a constant net force of 350 N is exerted on the weights, what is the net work done on the weight? (7.0 x 10 2 J) 3. A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 0 downward from the horizontal. Find the work done by the shopper on the cart as the shopper moves along a 50.0 m length of aisle. (1.6 x 10 3 J) 4. If 2.0 J of work is done in raising a 180 g apple, how gar is it lifted? (1.1 m)

The sign of work is important. Work is a scalar quantity and can be positive or negative. Work is positive when the component of force is in the same direction as the displacement. Work is negative when the force is in the direction opposite the displacement.

Homework In your notebook, write and answer the section review questions (1-6) on page 171 of your Physics textbook.